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Given below are two finite state automata ( $\rightarrow$ indicates the start state and $F$ indicates a final state)

$$\overset{Y}{\begin{array}{|l|l|l|}\hline \text{} & \textbf{a} & \textbf{b} \\\hline \text{\rightarrow 1} & \text{1} & \text{2} \\\hline \text{2 (F)} & \text{2} & \text{1} \\\hline \end{array}} \qquad \overset{Z}{\begin{array}{|l|l|l|}\hline \text{} & \textbf{a} & \textbf{b} \\\hline \text{\rightarrow 1} & \text{2} & \text{2} \\\hline \text{2 (F)} & \text{1} & \text{1} \\\hline \end{array}}$$

Which of the following represents the product automaton $Z \times Y$?

1. $\begin{array}{|l|l|}\hline \text{} & \text{a} & \text{b} \\\hline \text{$\rightarrowP$} & \text{S} & \text{R} \\\hline \text{Q} & \text{R} & \text{S} \\\hline \text{R(F)} & \text{Q} & \text{P}\\\hline \text{S} & \text{Q} & \text{P}\\\hline \end{array}$
2. $\begin{array}{|l|l|}\hline \text{} & \text{a} & \text{b} \\\hline \text{$\rightarrowP$} & \text{S} & \text{Q} \\\hline \text{Q} & \text{R} & \text{S} \\\hline \text{R(F)} & \text{Q} & \text{P}\\\hline \text{S} & \text{P} & \text{Q}\\\hline \end{array}$
3. $\begin{array}{|l|l|}\hline \text{} & \text{a} & \text{b} \\\hline \text{$\rightarrowP$} & \text{Q} & \text{S} \\\hline \text{Q} & \text{R} & \text{S} \\\hline \text{R(F)} & \text{Q} & \text{P}\\\hline \text{S} & \text{Q} & \text{P}\\\hline \end{array}$
4. $\begin{array}{|l|l|}\hline \text{} & \text{a} & \text{b} \\\hline \text{$\rightarrowP$} & \text{S} & \text{Q} \\\hline \text{Q} & \text{S} & \text{R} \\\hline \text{R(F)} & \text{Q} & \text{P}\\\hline \text{S} & \text{Q} & \text{P}\\\hline \end{array}$

I am getting answer as option A. but in last row instead of S -> Q | P I am getting S-> P | Q.
You are correct. That must be a printing mistake.
thanks
ans will be option B bcz in last two row entries are (Q,P) and (P,Q)

For those finding it difficult to map the states, in ZxY:

Start State → b→ Final State & Final State→ b→ Start State. This can be used to eliminate the options B, C, D. But although A satisfies first three transitions, last one wrongly given in the option.

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$$\begin{array}{|l|l|l|l|}\hline \textbf{} & \textbf{States} & \textbf{a} & \textbf{b} \\\hline \text{\rightarrow} & \textbf{11(P)} & 12 & 22 \\\hline \text{} & \textbf{12(S)} & 11 & 21 \\\hline \text{} & \textbf{21(Q)} & 22 & 12 \\\hline \textbf{(F)} & \textbf{22(R)} & 21 & 11 \\\hline \end{array}$$

$11$ is $P$ and $22$ is $R$ in choice. So, the answer should be (A) but in the row for $S$, it should be $P$ and $Q$ and not $Q$ and $P$.

by

Is there any difference in the product automaton of Z x Y and Y x Z ?
Does ZxY means concatenation of automata?
No, it is intersection
It is product rt?
Yes. it is , but when we are marking New (F) where finals of both FA's are together, then it results in Intersection in terms of language.
okay.. Let me check..
@Arjun sir ,  this is concatenation operation and you applied cross product method.. So, can we apply cross product method on concatenation operation also ??

@Arjun SIR , How did you know that P is 11 ?

I took P(11) Q(12)R(21)S(22) . Will it make any difference ? Is there any restriction to choose the states ?

In the end, try to map whichever is matching the given choices. No, other rule.

Arjun sir, I guess you have found Y x Z. Is Y x Z and Z x Y the same ?

@Arjun Sir, so can we consider that as a misprint or a typing error?
sir can you plz explain difference b/w Y x Z and Z x Y bcoz i think it is Y x Z. not Z x Y ??????
@rajoramanoj  Sir has actually shown it for YXZ   but anyway YXZ & ZXY are actually same, only order of state differs, both are having exactly same dfa if you check.

{1,1} = starting state =P   (given)

{1,2} = Q / S

{2,1} = Q/S

{2,2}=Final state = R  (given)

now correlate with options you will get your answer.

@Arjun-Sir your table seems to be for $Y \times Z$. But question is for $Z \times Y$

I got below table

 $Z \times Y$ a b 11 21 22 12 22 21 21 11 12 22 12 11

Is this correct?

I got the same. ( Answer will remain same i.e. option A for Z×Y and Y×Z)

@Harshada-Which table you got? Mine or That given in the answer by Arjun sir?

Same as your table @Ayush ! (Z×Y)

You could try NPTEL lectures by Kamala Kritivasan. She provides examples of DFA operations with state tables.

$Y:$

$Z:$

$Z\times Y:$

$P=11\ \&\ R=22$

$P\rightarrow b=R\ \&\ R\rightarrow b=P$ $\left(Only\ satisfied\ by\ option\ A\right)$

@ayush upadhaya sir mera bhi same table aya ha
@arjun sir i think you have made table of y*z which is incorrect

### Correct answer is option A.

New final state where finals of both FA's are together.

I don't  know even how to do it directly with Transition Table but with Figures, it is very easy.

for avoid confusion rename the states as 1,2,3,4.

and no doubt X means intersection.

In option A, S on a goes to Q

Hi,

When we do Z x Y then from Z → a (it goes to state 2), From Y → a (it goes to state 1), from 11 by reading a, it as to go 21, but your transition goes to 12 how?

let me know if I’m wrong?
@jeeruajay  You are correct I also got the same.

Another alternative to get the answer can be :

Y represents strings with odd number of b {Nb(W) mod 2 = 1)} and Z represents odd number of strings {|W| mod 2 =1}

If we take the product automata ZxY i.e. Odd number of String and Odd number of b in string which is nothing but "Strings with Odd no of b and Even no of a" Draw the mod m/c for this and pick the correct option i.e. A

### 1 comment

Yes this way we can check the correct option.

### 1 comment

@Mostafize Mondal

Bro you did Y X Z. But we have to perform

Z X Y.

But yeah we have to choose option A) ( last two rows are wrong-print mistake)